Question

PLZ HELP!
http://openstudy.com/study#/updates/5182ad9ce4b0e8530d3061a4

Answer 1

the equation i was given is
\[P(x)=_{n}C _{x}P ^{x}(1-P)^{n-x}\]

Answer 2

so go on the equation is correct.
what is the difficulty?

Answer 3

Well in the example i was given it goes on to \[\frac{ 11! }{ 0!(11-0)! }(0.35)^{0}(0.65)^{11}\] now i don't understand how to solve that.

Answer 4

yep thats correct. 0 factorial is 1 and anything raise to zero is also 1 so you'll get 1st term as 1*1*0.65^11

Answer 5

\[\frac{ 11! }{ 0!(11-0)! }(0.35)^0(0.65)^11\]

0!=1
(0.35)^0=1

\[\frac{ 11! }{ 11! }(1)(0.65)^{11}\]

\[(1)(1)(0.65)^11\]

Answer 6

(0.65)^11

0.00875

Answer 7

i put that into mathway and it gave me this. I KNOW that can't possibly be right. lol

Answer 8

it won't let me do that

Answer 9

so P(0)=0.00875 ?

Answer 10

thats why the answer is that much big value. just check for the factorial sign. and as you have to find probability for x<=4 just find probabilities for x=0,1,2,3,4 with the same formula and add them up. it 'll give you the probability for x<=4

Answer 11

the values will be:

0: 1x0.65^11

1: 11x0.65^10x0.35

2: (11x10/2x1)x 0.65^9x0.35^2

3: (11x10x9/3x2x1) x0.65^8 x 0.35^3

4: (11x10x9x8/4x3x2x1)x 0.65^7 x 0.35^4

calcualte each and sum them all, you'll be getting the answer: hope you understood this :D

Answer 12

ok so now i do \[P(1)=_{11}C _{1}(0.35)^{1}(1-0.35)^{11-1}\]
\[\frac{ 11! }{ 1!(11-1)! }(0.35)^{1}(0.65)^{10}\]

Answer 13

1!=1
0.35^1=0.35
now what do i do?

Answer 14

@sriramkumar
So P(0)=0.0963
P(1)=0.0518
Right?
i don't understand what the rest are even supposed to look like.

Answer 15

carry on the same for others too @joyce153 till you've reached p(4) then add them all.

Answer 16

@sriramkumar How u wrote (11x10/2x1)x 0.65^9x0.35^2
isn't making sense tho. I don't know how i'm supposed to write that out. The first 2 P(0) & P(1) made sense when writing it out but the rest don't.

Answer 17

its just 11C2 , 11C3 and 11C4 written in their simplified forms. when you cancel the like terms in the nCr you'll get the same. also its like a simple trick :D ;)

Answer 18

@joyce153 understood that???

Answer 19

not completely. i'm stuck on \[\frac{ 11! }{ 2!(11-2)! }(0.35)^2(0.65)^{11-2}\] can you explain it step by step on how to solve it cause it's like it all makes sense till i get to P(2) then i'm lost again.

Answer 20

its like 11! = 11 x 10! = 11x10x9!

and you have (11-2)! = 9! on the denominator cancel them? you'll get

11x10/2! = 11x10/2x1 ! that was the trick I used to write those all. now try the same! :D

Answer 21

understood??? @joyce153

Answer 22

one sec i'm trying to wrap my head around that. lol this is such a pain!

Answer 23

so is P(2)= 0.1395?

Answer 24

\[P(x \le4)=0.7558\] right? @sriramkumar

Answer 25

never mind it says its wrong. :(

Answer 26

But I'm sure the process is correct absolutely, there might be some calculation typo I think :(